Question: Simplify and expand the following expression: $ \dfrac{4}{4n + 28}+ \dfrac{2}{5n + 45}- \dfrac{4}{n^2 + 16n + 63} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{4}{4n + 28} = \dfrac{4}{4(n + 7)}$ We can factor a $5$ out of denominator in the second term: $ \dfrac{2}{5n + 45} = \dfrac{2}{5(n + 9)}$ We can factor the quadratic in the third term: $ \dfrac{4}{n^2 + 16n + 63} = \dfrac{4}{(n + 7)(n + 9)}$ Now we have: $ \dfrac{4}{4(n + 7)}+ \dfrac{2}{5(n + 9)}- \dfrac{4}{(n + 7)(n + 9)} $ The least common multiple of the denominators is: $ 20(n + 7)(n + 9)$ In order to get the first term over $20(n + 7)(n + 9)$ , multiply by $\dfrac{5(n + 9)}{5(n + 9)}$ $ \dfrac{4}{4(n + 7)} \times \dfrac{5(n + 9)}{5(n + 9)} = \dfrac{20(n + 9)}{20(n + 7)(n + 9)} $ In order to get the second term over $20(n + 7)(n + 9)$ , multiply by $\dfrac{4(n + 7)}{4(n + 7)}$ $ \dfrac{2}{5(n + 9)} \times \dfrac{4(n + 7)}{4(n + 7)} = \dfrac{8(n + 7)}{20(n + 7)(n + 9)} $ In order to get the third term over $20(n + 7)(n + 9)$ , multiply by $\dfrac{20}{20}$ $ \dfrac{4}{(n + 7)(n + 9)} \times \dfrac{20}{20} = \dfrac{80}{20(n + 7)(n + 9)} $ Now we have: $ \dfrac{20(n + 9)}{20(n + 7)(n + 9)} + \dfrac{8(n + 7)}{20(n + 7)(n + 9)} - \dfrac{80}{20(n + 7)(n + 9)} $ $ = \dfrac{ 20(n + 9) + 8(n + 7) - 80} {20(n + 7)(n + 9)} $ Expand: $ = \dfrac{20n + 180 + 8n + 56 - 80}{20n^2 + 320n + 1260} $ $ = \dfrac{28n + 156}{20n^2 + 320n + 1260}$ Simplify: $ = \dfrac{7n + 39}{5n^2 + 80n + 315}$